Question Library

(a) \(3y^{2} - 5y + 2 = 0\) \(\frac{3y^{2}}{3} - \frac{5y}{3} = \frac{-2}{3}\) \(y^{2} - \frac{5}{3}y = \frac{-2}{3}\) \(\frac{1}{2} of \frac{-5}{3} = \frac{-5}{6}\) \(y^{2} - \frac{5}{3}y + (-\frac{5}{6})^{2} = \frac{-2}{3} + (\frac{-5}{6})^{2}\) \((y - (\frac{5}{6}))^{2} = \frac{1}{36}\) Taking square root of both sides, \(\sqrt{(y - (\frac{5}{6}))^{2}} = \sqrt{\frac{1}{36}}\) \(y - \frac{5}{6} =\pm {\frac{1}{6}}\) \(y = \frac{5}{6} \pm \frac{1}{6}\) \(y = \frac{5}{6} + \frac{1}{6} = 1.00\) or \(y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3} = 0.67\). (b) \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\) \(MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 2n & 4x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\) \(m + 2n = 2 ... (1)\) \(4m + 2n = 3 ... (2)\) Multiplying (1) by 4, we have \(4m + 8n = 8 ... (3)\) (3) - (1) : \(8n - 2n = 8 - 2 = 6\) \(6n = 6 \implies n = 1\) \(m + 2n = m + 2(1) = m + 2 = 2\) \(m = 2 - 2 = 0\) \(x + 2y = 1 ... (1)\) \(4x + y = 4 ... (2)\) Multiply (1) by 4, \(4x + 8y = 4 ... (3)\) (3) - (2) : \(8y - y = 4 - 4 = 0\) \(\implies y = 0\) \(x + 2y = x + 2(0) = x + 0 = 1\) \(x = 1 - 0 = 1\) \(\therefore N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

3,9,27,81,..... r = \(\frac{9}{8}\) = 3 Tn = arn-1 Tn = 3(3)n-1 = \(\frac{3(3^2)}{3}\)

d \(\alpha\) t2 d = t2 k where k is a constant. d = 45cm, when t = 3s; thus 45 = 32 x t k = \(\frac{45}{9}\) = 5 thus equation connecting d and t is d = 5t2 when t = 6s, d = 5 x 62 = 5 x 36 = 180cm

Let x = \(\frac{4}{3}\), x = -\(\frac{3}{7}\) Then 3x = 4, 7x = -3 3x - 4 = 0, 7x + 3 = 0 (3x - 4)(7x + 3) = 0 21x2 + 9x - 28x - 12 = 0 21x2 - 19x - 12 = 0

0, 1, 1, 2, 3, 5, 8. 0 + 1 = 1 1+ 1 = 2 1+ 2 = 3 2 + 3 = 5 3 + 5 = 8 5 + 8 = 13 8 + 13 = 21 13 + 21 = 34

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ 240^\circ 270^\circ 300^\circ 330^\circ 360^\circ y 2.0 3.2 3.6 3.0 1.6 -0.2 -2.0 -3.2 -3.6 -3.0 -1.6 0.2 2.0 (b) (c)(i) The equation, \(3\sin x + 2\cos x = 0\) has solution where the curve cuts the x- axis, i.e. at A(x = 147^\circ) and B(x = 325.5^\circ). (ii)First, rearrange \(2 + 2\cos x + 3\sin x = 0\) to have the main graph content \(3\sin x + 2\cos x\) on one side of the equation; i.e. \(3\sin x + 2\cos x = -2\). Add the line \(y = -2\) to the graph. The line intersects the curve at the points (x = 180^\circ) and (x = 292.5^\circ). Hence, these are the solutions.

Starting from 3x − z \(\leq\) 9 x = 0, substitute the value of x (i.e. x = 0) into the equation 3x − z \(\leq\) 9 3(0) − z \(\leq\) 9 z \(\leq\) − 9 If z = 0 Then, 3x − 0 \(\leq\) 9 3x \(\leq\) 9 x \(\leq\) 3 Intercept of x and z i.e (x,z) = (3, -9)

Using the quadratic formula, we have \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5. \(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\) = \(\frac{4 \pm \sqrt{16 + 60}}{6}\) = \(\frac{4 \pm \sqrt{76}}{6}\) = \(\frac{4 \pm 8.72}{6}\) = \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\) = \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\) \(x = \text{2.12 or -0.79}\)

nth term of a linear sequence (AP) = a+(n − 1)d first term = 6, last term = 10 sum − 40 i.e. a = 6, l = 10, S = 40 \(S_n\)= n/2(2a + (n − 1)d or Sn = \div2 (a + l) \(S_n\) = n/2(a + l) 40 = n/2(6 + 10) 40 = 8n 8n = 40 8n = 40 n = 40/8 = 5 The number of terms = 5

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\) = \(\frac{4}{3}\) Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept. To get the value of b, we use a point on the line, say (0, 0). \(y = \frac{4}{3} x + b\) \(0 = \frac{4}{3}(0) + b\) \(b = 0\) The equation of the line is \(y = \frac{4}{3} x\)

No expanded working has been attached for this question yet. The verified answer is Option A.

1st G.P. = a =2 2nd G.P. = ar\(^{2 −1}\) = 54 2(r) = 54 r = 54/2 = 27 r = 27 3rd term = ar\(^2\) = (2) (27)\(^2\) 2 \times 27 \times 27 = 1458

\(27^x ÷ 81^{(x + 2y)} = 9 \\ (27)x = 9 × 81^{(x+2y)} \\ (3^3 )^x =32 \times 3^{4(x + 2y)} \\ =3^{(2 + 4x + 8y)}\\ 3^{3x} = 3^{ (2 + 4x + 8y)}\\ 3x = 2 + 4x + 8y\\ 3x − 4x − 8y = 2 … … … (1)\\ x + 4y = 0 … … … (2)\\ − 4y = 2\\ y = (− 2) ÷ 4 = − ½\\ y = − ½\\ \) Substitute the value of y into equation (2) i.e x + 4y = 0 x + 4( − 1/2) = 0 x − 2 = 0 x = 2 ∴ x = 2,y = − ½) Method II \( 27^x ÷ 31^{(x + 2y) }= 9\\ 3^{3x} × 3^{( − 4x − 8y)} = 32\\ 3^{(3x − 8y)} = 32\\ − x − 8y=2 ……… (1)\\ x + 4y = 0 ……… (2)\\ − 4 = 2\\ y= 2/4 = ½\\ y = ½ \) Substitute the value of y into equation 2 x + 4y=0 x + 4 (− 1) ÷ 2) = 0 x − 2 = 0 x = 2 x = 2, y = ½

\( 5^{(x − 2)} = (125^{−1} )^{(x + 3)}\\ 5^{(x − 2)} = (5− 3)^{x + 3}\\ 5^{(x − 2)} = 5^{− 3x − 9}\\ x + 3x = − 9 + 2\\ 4x = − 7\\ x = \frac{−7}{4} \)

GIVEN: 2, 5, 10, 17, 26,. The numbers are increasing by odd value. 1st term = 2 2nd term = 5 3rd term = 10 4th term = 17 5th term = 26 6th term = 37 7th term = 50 Always add 2 to the difference between the two preceding numbers and add this to the later number to get the next term.